Question

The Derivative of x^x^x ... ! Thanks !! =)

Answer 1

x^(-1+x+x^x) (1+x log(x)+x log^2(x))

Answer 2

What ? how did get the answer ? i don't even know how to start !! !

Answer 3

if you are familiar with the chain rule then it isnt too bad

Answer 4

i am .. but can you give me the guidelines at least for starting the process of the calculation ? =)

Answer 5

first, let \[y = x^{x^x}\]
take the ln of both sides
\[\ln y = x\ln x^x\]
take the derivative of both sides
\[\frac 1y y' = \ln x^x + x \frac{d}{dx} (\ln x^x)\]
side solution:
\[\frac{d}{dx} (\ln x^x) \implies \frac{\frac{d}{dx} x^x}{x^x}\]
so what is derivative of x^x?
let
\[a = x^x\]
\[\ln a = x\ln x\]
\[\frac 1a a' = 1 + \ln x\]
\[a' = x^x(1 + \ln x)\]
so that's the derivative of x^x

\[\implies \frac 1y y' = \frac{x^x(1 + \ln x)}{x^x}\]
\[\implies y' = y[\frac{x^x (1 + \ln x)}{x^x}]\]
\[\implies y' = x^{x^x} [\frac{x^x(1 + \ln x)}{x^x}]\]
does that help?

Answer 6

lol i missed a step

Answer 7

Wow thanks guys !!! you are very helpful !! wish i could give you both medals !!!! =\

Answer 8

i missed a step there...
\[\frac 1y y' = \ln x^x + x[ \frac{x^x ( 1 + \ln x)}{x^x}]\]
that's how it should have been

Answer 9

Thank you very much !!!!!!